0000:00:00:00:00Start4Number 4ExplanationIf (p,q)(p, q)(p,q) is the vertex point of the graph of function f(x)=ax2+2ax+a+1f(x) = ax^2 + 2ax + a + 1f(x)=ax2+2ax+a+1, with f(a)=19f(a) = 19f(a)=19, then p+2q+3a=....p + 2q + 3a = ....p+2q+3a=....777666000−1-1−1−2-2−2ExplanationGiven f(a)=19f(a) = 19f(a)=19 then f(x)=ax2+2ax+a+1f(x) = ax^2 + 2ax + a + 1f(x)=ax2+2ax+a+1f(a)=19f(a) = 19f(a)=19a⋅a2+2a⋅a+a+1=19a \cdot a^2 + 2a \cdot a + a + 1 = 19a⋅a2+2a⋅a+a+1=19a3+2a2+a−18=0a^3 + 2a^2 + a - 18 = 0a3+2a2+a−18=0 Factor using Horner's method to get a3+2a2+a−18=0a^3 + 2a^2 + a - 18 = 0a3+2a2+a−18=0(a−2)(a2+4a+9)=0(a - 2)(a^2 + 4a + 9) = 0(a−2)(a2+4a+9)=0a−2=0∪(a2+4a+9)=0a - 2 = 0 \cup (a^2 + 4a + 9) = 0a−2=0∪(a2+4a+9)=0a=2∪a2+4a+9=0a = 2 \cup a^2 + 4a + 9 = 0a=2∪a2+4a+9=0 Thus we obtain a=2a = 2a=2, and the quadratic function becomes f(x)=ax2+2ax+a+1f(x) = ax^2 + 2ax + a + 1f(x)=ax2+2ax+a+1f(x)=2x2+2⋅2x+2+1f(x) = 2x^2 + 2 \cdot 2x + 2 + 1f(x)=2x2+2⋅2x+2+1f(x)=2x2+4x+3f(x) = 2x^2 + 4x + 3f(x)=2x2+4x+3 Vertex point ppp p=−b2a=−42(2)=−1p = \frac{-b}{2a} = \frac{-4}{2(2)} = -1p=2a−b=2(2)−4=−1q=f(p)=f(−1)q = f(p) = f(-1)q=f(p)=f(−1)=2(−1)2+4(−1)+3= 2(-1)^2 + 4(-1) + 3=2(−1)2+4(−1)+3q=1q = 1q=1 Therefore the value of p+2q+3ap + 2q + 3ap+2q+3a is −1+2(1)+3(2)=−1+2+6=7-1 + 2(1) + 3(2) = -1 + 2 + 6 = 7−1+2(1)+3(2)=−1+2+6=7
4Number 4ExplanationIf (p,q)(p, q)(p,q) is the vertex point of the graph of function f(x)=ax2+2ax+a+1f(x) = ax^2 + 2ax + a + 1f(x)=ax2+2ax+a+1, with f(a)=19f(a) = 19f(a)=19, then p+2q+3a=....p + 2q + 3a = ....p+2q+3a=....777666000−1-1−1−2-2−2ExplanationGiven f(a)=19f(a) = 19f(a)=19 then f(x)=ax2+2ax+a+1f(x) = ax^2 + 2ax + a + 1f(x)=ax2+2ax+a+1f(a)=19f(a) = 19f(a)=19a⋅a2+2a⋅a+a+1=19a \cdot a^2 + 2a \cdot a + a + 1 = 19a⋅a2+2a⋅a+a+1=19a3+2a2+a−18=0a^3 + 2a^2 + a - 18 = 0a3+2a2+a−18=0 Factor using Horner's method to get a3+2a2+a−18=0a^3 + 2a^2 + a - 18 = 0a3+2a2+a−18=0(a−2)(a2+4a+9)=0(a - 2)(a^2 + 4a + 9) = 0(a−2)(a2+4a+9)=0a−2=0∪(a2+4a+9)=0a - 2 = 0 \cup (a^2 + 4a + 9) = 0a−2=0∪(a2+4a+9)=0a=2∪a2+4a+9=0a = 2 \cup a^2 + 4a + 9 = 0a=2∪a2+4a+9=0 Thus we obtain a=2a = 2a=2, and the quadratic function becomes f(x)=ax2+2ax+a+1f(x) = ax^2 + 2ax + a + 1f(x)=ax2+2ax+a+1f(x)=2x2+2⋅2x+2+1f(x) = 2x^2 + 2 \cdot 2x + 2 + 1f(x)=2x2+2⋅2x+2+1f(x)=2x2+4x+3f(x) = 2x^2 + 4x + 3f(x)=2x2+4x+3 Vertex point ppp p=−b2a=−42(2)=−1p = \frac{-b}{2a} = \frac{-4}{2(2)} = -1p=2a−b=2(2)−4=−1q=f(p)=f(−1)q = f(p) = f(-1)q=f(p)=f(−1)=2(−1)2+4(−1)+3= 2(-1)^2 + 4(-1) + 3=2(−1)2+4(−1)+3q=1q = 1q=1 Therefore the value of p+2q+3ap + 2q + 3ap+2q+3a is −1+2(1)+3(2)=−1+2+6=7-1 + 2(1) + 3(2) = -1 + 2 + 6 = 7−1+2(1)+3(2)=−1+2+6=7
